Question

find the value of k such that 3 X square + 2 k x + X - K - 5 has sum of zeroes as half of their product ?

Answer 1

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Let p( x ) = 3x² + 2kx + x - k - 5

= 3x² + ( 2k + 1 ) x - ( k + 5 )

compare p( x ) with ax² + bx + c

a = 3 , b = 2k + 1 , c = - ( k + 5 )

1 ) sum of the zeroes = - b / a

= - ( 2k + 1 ) / 3

2 ) product of the zeroes = c /a

= - ( k + 5 ) / 3

according to the problem given ,

- ( 2k + 1 ) / 3 = 1/2 [ - ( k + 5 ) / 3 ]

2 ( 2k + 1 ) = ( k + 5 )

4k + 2 = k + 5

4k - k = 5 - 2

3k = 3

k = 3/3

k = 1

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Answer 2

Answer:

Step-by-step explanation:

Hi ,

Let p( x ) = 3x² + 2kx + x - k - 5

= 3x² + ( 2k + 1 ) x - ( k + 5 )

compare p( x ) with ax² + bx + c

a = 3 , b = 2k + 1 , c = - ( k + 5 )

1 ) sum of the zeroes = - b / a

= - ( 2k + 1 ) / 3

2 ) product of the zeroes = c /a

= - ( k + 5 ) / 3

according to the problem given ,

- ( 2k + 1 ) / 3 = 1/2 [ - ( k + 5 ) / 3 ]

2 ( 2k + 1 ) = ( k + 5 )

4k + 2 = k + 5

4k - k = 5 - 2

3k = 3

k = 3/3

k = 1

I hope this helps you.

:)